The ion-product constant for water is \(1.0 \times 10^{-14}\) at \(25^{\circ} \mathrm{C}\) and \(3.8 \times 10^{-14}\) at \(40^{\circ} \mathrm{C} .\) Is the forward process $$ \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ endothermic or exothermic?

Short Answer

Expert verified
The forward process \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)\) is endothermic.

Step by step solution

01

Recognize the Principle

Recognize and apply Le Châtelier's principle to analyze this exercise. The principle can be summarized as: If an external condition of a reaction in equilibrium changes, the system will adjust itself to partially counteract the change.
02

Compare the Ion-Product Constants

Compare the ion-product constant for water at \(25^{\circ} \mathrm{C}\) which is \(1.0 \times 10^{-14}\), and at \(40^{\circ} \mathrm{C}\) which is \(3.8 \times 10^{-14}\). Observe that the ion-product constant has increased as the temperature increases.
03

Analyze the Change in Equilibrium Position

Since it has been observed that the ion-product constant increases with an increase in temperature, this would imply that the equilibrium position has shifted to the right, favoring the forward process. As the forward reaction is favored, this suggests that the heat energy is absorbed by the system from the surroundings.
04

Identify the Type of Reaction

The heat energy absorbed by the system is a characteristic of an endothermic reaction. Hence, the forward process \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)\) is endothermic

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